$x$, $y$, and $\frac3{2x}$ are non-zero terms in an arithmetic progression. If the third term is increased by $1$, the three terms now form a geometric progression. Find the original three terms.

Here are my steps:

1) The numbers were set into a proportion to find the common ratio. $\dfrac xy = \dfrac{\tfrac3{2x}}y$

2) The equation was set equal to $0$ such that $0=\tfrac32x^2-y^2$

I do not know how to move on from here or if the steps lead to the correct answer. How would you solve this problem?

  • $\begingroup$ It's wrong. What you are doing is treating the original sequence as a geometric progression but what the question says is the original sequence is arithmetic. $\endgroup$ – cr001 Feb 11 at 11:46
  • $\begingroup$ @cr001 In that case, do I find the common difference? $\endgroup$ – Bibliophile Feb 11 at 12:01
  • 1
    $\begingroup$ Yes, that's correct. $\endgroup$ – cr001 Feb 11 at 12:03

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